Timeline for Why variance of OLS estimate decreases as sample size increases?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2024 at 2:24 | comment | added | mathdoge | For those who don't like the Woodbury identity, you could do a simultaneous diagonalization to $X^T X$ and $x^T x$, because they are symmetric and PSD. | |
| Dec 28, 2021 at 22:47 | comment | added | AAL | There was indeed a typo in the denominator: missing a factor of $(X'X)^{-1}$. I have edited the answer accordingly and checked that it fixes the counterexample. Thank you for pointing it out. | |
| Dec 28, 2021 at 22:35 | history | edited | AAL | CC BY-SA 4.0 | Corrected typo in equation |
| Dec 28, 2021 at 18:36 | comment | added | whuber♦ | A careful reader, without enough rep to comment, has called to my attention an error in the application of this identity. Indeed, the denominator of the right hand side of the equality cannot be correct because it does not scale properly upon a simple rescaling of the variables. (Alternatively, consider $X=(1,1)^\prime$ (for $p=0$) and $x=(1).$ It asserts $(2)^{-1}=(2)^{-1} - (2)^{-1}(1)(2)^{-1}/(1+1),$ equivalent to $1/3=3/8.$) The argument is a good one, so it would be nice to get the details right. stats.stackexchange.com/a/90921/919 shows what the formula ought to look like. | |
| Jun 24, 2020 at 16:28 | review | Late answers | |||
| Jun 24, 2020 at 17:03 | |||||
| Jun 24, 2020 at 16:22 | comment | added | whuber♦ | +1. Thank you for the clear explanation -- and welcome to CV. | |
| Jun 24, 2020 at 16:18 | review | First posts | |||
| Jun 24, 2020 at 18:38 | |||||
| Jun 24, 2020 at 16:11 | history | answered | AAL | CC BY-SA 4.0 |